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More Functions and Sets Rosen 1.8

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Inverse Image Let f be an invertible function from set A to set B. Let S be a subset of B. We define the inverse image of S to be the subset of A containing all pre-images of all elements of S. f -1 (S) = {a A | f(a) S} S A f(a) a B

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Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S) What do we know? f must be 1-to-1 and onto S A B b1a2b2 a1 ffff f -1 (S)

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Proof: We must show that f -1 (S) f -1 (S) and that f -1 (S) f -1 (S). Let x f -1 (S). Then x A and f(x) S. Since f(x) S, x f -1 (S). Therefore x f -1 (S). Now let x f -1 (S). Then x f -1 (S) which implies that f(x) S. Therefore f(x) S and x f -1 (S) Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S)

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Proof: f -1 (S) = {x A | f(x) S}Set builder notation = {x A | f(x) S}Def of Complement = f -1 (S)Def of Complement Let f be an invertible function from A to B. Let S be a subset of B. Show that f -1 (S) = f -1 (S)

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Floor and Ceiling Functions The floor function assigns to the real number x the largest integer that is less than or equal to x. x x = n iff n x < n+1, n Z x = n iff x-1 < n x, n Z The ceiling function assigns to the real number x the smallest integer that is greater than or equal to x. x x = n iff n-1 < x n, n Z x = n iff x n < x+1, n Z

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Examples 0.5 = 1 0.5 = 0 -0.3 = 0 -0.3 = -1 6 = 6 6 = 6 -3.4 = -3 3.9 = 3

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Prove that x+m = x + m when m is an integer. Proof: Assume that x = n, n Z. Therefore n x < n+1. Next we add m to each term in the inequality to get n+m x+m < n+m+1. Therefore x+m = n+m = x + m x = n iff n x < n+1, n Z

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Let x R. Show that 2x = x + x+1/2 Proof: Let n Z such that x = n. Therefore n x < n+1. We will look at the two cases: x n + 1/2 and x < n + 1/2. Case 1: x n + 1/2 Then 2n+1 2x < 2n+2, so 2x = 2n+1 Also n+1 x + 1/2 < n+2, so x + 1/2 = n+1 2x = 2n+1 = n + n+1 = x + x+1/2 nn+1

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Let x R. Show that 2x = x + x+1/2 Case 2: x < n + 1/2 Then 2n 2x < 2n+1, so 2x = 2n Also n x + 1/2 < n+1, so x + 1/2 = n 2x = 2n = n + n = x + x+1/2

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Characteristic Function Let S be a subset of a universal set U. The characteristic function f S of S is the function from U to {0,1}such that f S (x) = 1 if x S and f S (x) = 0 if x S. Example: Let U = Z and S = {2,4,6,8}. f S (4) = 1 f S (10) = 0

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Let A and B be sets. Show that for all x, f A B (x) = f A (x)f B (x) Proof: f A B (x) must equal either 0 or 1. Suppose that f A B (x) = 1. Then x must be in the intersection of A and B. Since x A B, then x A and x B. Since x A, f A (x)=1 and since x B f B (x) = 1. Therefore f A B = f A (x)f B (x) = 1. If f A B (x) = 0. Then x A B. Since x is not in the intersection of A and B, either x A or x B or x is not in either A or B. If x A, then f A (x)=0. If x B, then f B (x) = 0. In either case f A B = f A (x)f B (x) = 0.

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Let A and B be sets. Show that for all x, f A B (x) = f A (x) + f B (x) - f A (x)f B (x) Proof: f A B (x) must equal either 0 or 1. Suppose that f A B (x) = 1. Then x A or x B or x is in both A and B. If x is in one set but not the other, then f A (x) + f B (x) - f A (x)f B (x) = 1+0+(1)(0) = 1. If x is in both A and B, then f A (x) + f B (x) - f A (x)f B (x) = 1+1 – (1)(1) = 1. If f A B (x) = 0. Then x A and x B. Then f A (x) + f B (x) - f A (x)f B (x) = 0 + 0 – (0)(0) = 0.

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Let A and B be sets. Show that for all x, f A B (x) = f A (x) + f B (x) - f A (x)f B (x) ABA B f A B (x) f A (x) + f B (x) - f A (x)f B (x) 1 1 111+1-(1)(1) = 1 1 0 111+0-(1)(0) = 1 0 1 110+1-(0)(1) = 1 0 0 000+)-(0)(0) = 0

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